## Units of Partial Pressure in 5G-13, 5G-15

$\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$

Sofia Barker 2C
Posts: 101
Joined: Wed Sep 18, 2019 12:21 am

### Units of Partial Pressure in 5G-13, 5G-15

For 5G 13 and 5G 15, the textbook solutions manual keeps the product and reactant partial pressures in units of bar to calculate Q. Why isn't it necessary to convert to atmospheres? I calculated Q using bar and then again using atm and the values are different, so the reasoning can't be that they are equivalent or close to equivalent in value.

Vicki Liu 2L
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am

### Re: Units of Partial Pressure in 5G-13, 5G-15

Actually, I think 1 bar is approximately equal to 1 atm, so it should give you the same answer. Online it says 1 bar = 0.986923 atm.

Caitlyn Tran 2E
Posts: 100
Joined: Fri Aug 09, 2019 12:15 am

### Re: Units of Partial Pressure in 5G-13, 5G-15

As stated before, bars are approximately equal to atmospheres, but in general this shouldn't matter as long as they are all in the same units. When you are calculating Q, you usually use the values without units because the partial pressures are representing the activities of each compound, which has no units. Hope this helps!

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

### Re: Units of Partial Pressure in 5G-13, 5G-15

Calculating the Q value with partial pressures gives you an answer without units. Therefore, it isn't necessary to convert the values in bars.